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I'm encountered the calculation of this function [itex]\int^{\infty}_0 e^{-x^2} f(x) dx[/itex]. How to do it? Thanks.

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- Thread starter jollage
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- #1

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I'm encountered the calculation of this function [itex]\int^{\infty}_0 e^{-x^2} f(x) dx[/itex]. How to do it? Thanks.

- #2

SteamKing

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Depends of what f(x) is.

- #3

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Depends of what f(x) is.

SteamKing, thanks. I have another problem, which I encountered in the same paper. Maybe you could help me. Thanks in advance.

The author asserts that [itex]V(k,t)=V(k,0)e^{-k^2t}+ k\int^{t}_{0}C(t')e^{k^2(t'-t)}dt'[/itex] implies [itex]V(k,t)=C(t)/k + \mathcal{O}(k^{-3})[/itex] when t>0 and [itex]k\rightarrow \infty[/itex]. Could you see this?

- #4

mathman

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It looks like the V(k,0) term -> 0 very fast, so it can be ignored.

The lntegrand of the integral looks as if -> δ(t'-t)/k

- #5

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It looks like the V(k,0) term -> 0 very fast, so it can be ignored.

The lntegrand of the integral looks as if -> δ(t'-t)/k^{2}.

Mathman, thanks. I agree with you, but I need the detail to understand it.. Since it seems it is impossible to expand the exponential term in the integral, I don't know other techniques to evaluate the integral...

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